JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Conservation of Momentum

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A pendulum of length 2 m consists of a wooden bob of mass 50 g. A bullet of mass 75 g is fired towards the stationary bob with a speed v. The bullet emerges out of the bob with a speed

\frac{v}{3}

and the bob just completes the vertical circle. The value of

v

is ______

m s^{- 1}

. (if

g = 10 m/s^{2}

)

Answer: 10

Solution

For the bob to just complete a vertical circle of radius

R = 2

m, its velocity at the bottom must be

v_{bob} = 5 g R = 5 \times 10 \times 2 = 10

m/s.

Apply conservation of momentum in the horizontal direction during the collision:

m_{bullet} \cdot v = m_{bullet} \cdot \frac{v}{3} + m_{bob} \cdot v_{bob}

75 v = 75 \cdot \frac{v}{3} + 50 \cdot 10

75 v - 25 v = 500

50 v = 500 ⟹ v = 10 m/s

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer

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